Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(X1, X2)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(X1, X2)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK1(f2(X1, X2)) -> A__F2(mark1(X1), X2)
A__F2(g1(X), Y) -> MARK1(X)
A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
MARK1(g1(X)) -> MARK1(X)
MARK1(f2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

MARK1(f2(X1, X2)) -> A__F2(mark1(X1), X2)
A__F2(g1(X), Y) -> MARK1(X)
A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
MARK1(g1(X)) -> MARK1(X)
MARK1(f2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(f2(X1, X2)) -> A__F2(mark1(X1), X2)
A__F2(g1(X), Y) -> MARK1(X)
A__F2(g1(X), Y) -> A__F2(mark1(X), f2(g1(X), Y))
MARK1(g1(X)) -> MARK1(X)
MARK1(f2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK1(x1) ) = x1 + 2


POL( a__f2(x1, x2) ) = 2x1 + 1


POL( mark1(x1) ) = max{0, 2x1 - 1}


POL( A__F2(x1, x2) ) = x1 + 1


POL( g1(x1) ) = 2x1 + 2


POL( f2(x1, x2) ) = 2x1 + 1



The following usable rules [14] were oriented:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
a__f2(X1, X2) -> f2(X1, X2)
mark1(g1(X)) -> g1(mark1(X))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__f2(g1(X), Y) -> a__f2(mark1(X), f2(g1(X), Y))
mark1(f2(X1, X2)) -> a__f2(mark1(X1), X2)
mark1(g1(X)) -> g1(mark1(X))
a__f2(X1, X2) -> f2(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.